Question

Determine an approximate value of \(\phi(0.2)\), by the Euler method and the Runge-Kutta method, for the initial-value problem $$ \begin{array}{ll} x^{\prime}=x+4 y, & x(0)=1 \\ y^{\prime}=x-y, & y(0)=0 . \end{array} $$

Short answer

Based on the given system of ordinary differential equations (ODEs): $$ \begin{array}{ll} x^{\prime}=x+4y, & x(0)=1 \\\ y^{\prime}=x-y, & y(0)=0 \end{array} $$ We used Euler's method and the 4th order Runge-Kutta method to approximate the value of φ(0.2) (or y(0.2)). The results are as follows: - Euler's method: 0.21 - Runge-Kutta 4 method: 0.25833

Step-by-step solution

Identify the ODEs and initial conditions

We are given the following system of ODEs: $$ \begin{array}{ll} x^{\prime}=x+4y, & x(0)=1 \\\ y^{\prime}=x-y, & y(0)=0 \end{array} $$ We need to find the approximate value of φ(0.2), which is the same as finding the value of y(0.2) using our given initial values x(0) = 1 and y(0) = 0.

Implement the Euler method

Since Euler's method is a first-order method, let's first implement it to find an approximate value for y(0.2). For the Euler method, the update rule for each iteration is: $$ x_{i+1} = x_{i} + h(f_1(x, y)) \\ y_{i+1} = y_{i} + h(f_2(x, y)) $$ where f1(x, y) = x + 4y and f2(x, y) = x - y. Consider h = 0.1 for simplicity.

Calculate the Euler method values

Following the Euler method update rule for two iterations with h = 0.1: $$ x_1 = x_0 + 0.1(f_1(x_0, y_0)) = 1 + 0.1(1+4\times 0) = 1.1 \\ y_1 = y_0 + 0.1(f_2(x_0, y_0)) = 0 + 0.1(1-0) = 0.1 $$ $$ x_2 = x_1 + 0.1(f_1(x_1, y_1)) = 1.1 + 0.1(1.1+4\times 0.1) = 1.25 \\ y_2 = y_1 + 0.1(f_2(x_1, y_1)) = 0.1 + 0.1(1.1-0.1) = 0.21 $$ The Euler method approximation of φ(0.2) is y(0.2) = 0.21.

Implement the Runge-Kutta method

Now we will implement the 4th order Runge-Kutta method, which is more accurate compared to Euler's method. The update rule for the Runge-Kutta 4 method is as follows: $$ k_{1_1} = f_1(x, y) \\ k_{1_2} = f_2(x, y) \\ k_{2_1} = f_1(x + \frac{h}{2}, y + \frac{h}{2}k_{1_1}) \\ k_{2_2} = f_2(x + \frac{h}{2}, y + \frac{h}{2}k_{1_2}) \\ k_{3_1} = f_1(x + \frac{h}{2}, y + \frac{h}{2}k_{2_1}) \\ k_{3_2} = f_2(x + \frac{h}{2}, y + \frac{h}{2}k_{2_2}) \\ k_{4_1} = f_1(x + h, y + hk_{3_1}) \\ k_{4_2} = f_2(x + h, y + hk_{3_2}) \\ x_{i+1} = x_{i} + \frac{h}{6}(k_{1_1} + 2k_{2_1} + 2k_{3_1} + k_{4_1}) \\ y_{i+1} = y_{i} + \frac{h}{6}(k_{1_2} + 2k_{2_2} + 2k_{3_2} + k_{4_2}) $$

Calculate the Runge-Kutta method values

Following the Runge-Kutta 4 update rule for two iterations with h = 0.1: Iteration 1: $$ k_{1_1} = f_1(1,0) = 1 \\ k_{1_2} = f_2(1,0) = 1 \\ k_{2_1} = f_1(1.05, 0.05) = 1.2000 \\ k_{2_2} = f_2(1.05, 0.05) = 1 \\ k_{3_1} = f_1(1.05, 0.1) = 1.2000 \\ k_{3_2} = f_2(1.05, 0.1) = 0.9000 \\ k_{4_1} = f_1(1.1, 0.2) = 1.4 \\ k_{4_2} = f_2(1.1, 0.2) = 0.7000 $$ Update with the Runge-Kutta method values: $$ x_1 = 1 + \frac{1}{60}(1+4+4+1.4) = 1.15 \\ y_1 = 0 + \frac{1}{60}(1+4+4+0.7) = 0.150 $$ Iteration 2: $$ k_{1_1} = f_1(1.15,0.15) = 1.75 \\ k_{1_2} = f_2(1.15,0.15) = 1 \\ k_{2_1} = f_1(1.225,0.225) = 1.925 \\ k_{2_2} = f_2(1.225,0.225) = 0.925 \\ k_{3_1} = f_1(1.225,0.3) = 1.925 \\ k_{3_2} = f_2(1.225,0.3) = 0.825 \\ k_{4_1} = f_1(1.3,0.45) = 2.1 \\ k_{4_2} = f_2(1.3,0.45) = 0.550 $$ Update with the Runge-Kutta method values: $$ x_2 = 1.15 + \frac{1}{60}(1.75+3.85+3.85+2.1) = 1.31583 \\ y_2 = 0.15 + \frac{1}{60}(1+3.7+3.3+0.55) = 0.25833 $$ The Runge-Kutta method approximation of φ(0.2) is y(0.2) = 0.25833. In conclusion, the approximate value of φ(0.2), or y(0.2), is: - Euler's method: 0.21 - Runge-Kutta 4 method: 0.25833

Key concepts

Euler Method

Euler's method is a simple numerical technique used to find approximate solutions for ordinary differential equations (ODEs). It's particularly useful when an equation can't be solved analytically. Here's how it works:

1. **Starting Point:** Begin at the initial values given in the problem. For example, if the problem states
\( x(0) = 1 \) and \( y(0) = 0 \), you start your calculations here.
2. **Step Size (\( h \)):** Choose a small step size. Smaller steps give more accurate results. In this example, \( h = 0.1 \).
3. **Iterate:** Use the formulas \( x_{i+1} = x_i + h(f_1(x_i, y_i)) \) and \( y_{i+1} = y_i + h(f_2(x_i, y_i)) \) to calculate new values.

Repeat this process for the required number of steps to approximate the solution at a specific point. It's a straightforward method but can be less accurate, especially for larger steps.

Runge-Kutta Method

The Runge-Kutta method is an extension of Euler's method that offers greater accuracy by using intermediate steps to calculate the slope. The fourth-order Runge-Kutta (RK4) method is especially popular. Here’s why:

1. **Intermediate Calculations:** It computes several intermediate values \( k_1, k_2, k_3, \) and \( k_4 \) at different points, giving a more refined slope.
2. **Weighted Average:** Combines these intermediate results through a weighted average to find the final approximation for \( x \) and \( y \).
3. **Steps Involved:**
- Calculate \( k_1 \) using the initial point.
- Use \( k_1 \) to find \( k_2 \), and continue similarly for \( k_3 \) and \( k_4 \).
- Finally, find the next value of \( y \) using \( y_{i+1} = y_i + \frac{h}{6}(k_1 + 2k_2 + 2k_3 + k_4) \).

This technique is more complex but provides significantly better accuracy, making it ideal for more precise applications.

Initial Value Problem

An initial value problem consists of an ordinary differential equation paired with a specific starting value. This starting point, called the initial condition, dictates the solution's trajectory. Here's what you need to know:

1. **Definition:** Given a differential equation like \( x' = x + 4y \), along with initial conditions like \( x(0) = 1 \), it qualifies as an initial value problem.
2. **Purpose:** These problems aim to find a function that not only satisfies the differential equation but also meets the given condition at the starting point.
3. **Applications:** Common in physics and engineering, these problems model scenarios where conditions at a given time must evolve according to specific rules.

Knowing the initial conditions allows numerical methods to compute approximate solutions at other points, evolving step by step from the given start.

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) involve functions of one independent variable and their derivatives. Here's why they're essential:

1. **Types:** ODEs can be first-order, such as \( y' = f(x, y) \), or higher-order with more derivatives.
2. **Applications:** They model phenomena like motion, growth, decay, and more. Many scientific and engineering problems rely on ODEs.
3. **Solution Methods:** Analytically, some ODEs have solutions. However, for more complex cases, numerical methods like Euler and Runge-Kutta are used.

The essence of an ODE is to describe how a quantity changes with respect to another, forming the backbone of many dynamic models in various fields.